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On a graph of a system under a external force y = distance and x = time where the external force start at t = 0, it's easy to find the driving frequency.

$$F = \frac{\omega}{2\pi}, \omega = \frac{2\pi}{T}$$ and we can get $T$ easily with the steady state part of the graph.

However, is there a way to find the natural frequency?

Maybe by finding where $w_d = w_0$, which is the resonance frequency.

I made a graph. Can I consider steady-state amplitude as the driving force amplitude?

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  • $\begingroup$ Yes, you look for the resonance peak $\endgroup$ – FGSUZ Feb 7 at 19:35
  • $\begingroup$ I'm not sure what you mean by resonance peak. $\endgroup$ – proxima Feb 7 at 20:28
  • $\begingroup$ What is on the y axis of the graph? The displacement of the oscilaltor versus time, since the driving force starts to act? $\endgroup$ – nasu Feb 7 at 21:09
  • $\begingroup$ @nasu The system is in equilibrium at x=0. $\endgroup$ – proxima Feb 7 at 21:21
  • $\begingroup$ Driving force = input, steady state amplitude = output displacement. So no, driving force amplitude != steady state amplitude $\endgroup$ – oliver Feb 7 at 21:29
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If this is real data (or if you have real data) the first part of the graph may represent what is called the transient regime. In this regime the motion is a superposition of two motions, one with the natural frequency and another with the driving frequency. After some time the motion with natural frequency dies out. So, if you have enugh data for the transient regime to do a decent Fourier analysis you may be able to see two peaks in the spectrum, one for each of the two frequencies. But I am not sure it will work for this curve.

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